Shortest Confidence interval for $\sigma^2$ I have a $X_1, \cdots , X_n$ random samples from a $N(\mu,\sigma^2)$. $\mu$ is known, but $\sigma^2$ is  unknown. I would like to know how to go about constructing a $(1-\alpha)$100% shortest confidence interval for $\sigma^2$.  
I know how how to construct that of a known variance, but unknown mean. I'm bit confused as how to proceed when the situation is reversed.
Thank you.
 A: First note that $\mathbb{E}\left[(X-\mu)^2 \right] = \sigma^2$
An unbiased estimator of $\sigma^2$ is given by$$\hat{s}^2 = \displaystyle \frac{\displaystyle \sum_{k=1}^{n} \left( X_i - \mu \right)^2}{n}$$
Now we have that $\displaystyle \frac{n \hat{s}^2}{\sigma^2}$ is a $\chi^2_n$ i.e. a $\chi^2$ random variable with $n$ degrees of freedom (since $X_i$ are normally distributed random variables)
Find $a$ and $b$ such that $\displaystyle 1- \frac{\alpha}{2} = P\left(\chi_n^2 \leq b \right)$ and $\displaystyle \frac{\alpha}{2} = P\left(\chi_n^2 \leq a \right)$
Hence, we have $$\displaystyle P \left(a \leq \chi_n^2 \leq b \right) = 1 - \alpha$$
$$\displaystyle P \left(a \leq \frac{n \hat{s}^2}{\sigma^2} \leq b \right) = 1 - \alpha$$
$$\displaystyle P \left(\frac{n \hat{s}^2}{b} \leq \sigma^2 \leq \frac{n \hat{s}^2}{a} \right) = 1 - \alpha$$
Hence, the desired $1-\alpha$ confidence interval is given by
$$\left(\frac{n \hat{s}^2}{b},\frac{n \hat{s}^2}{a} \right)$$ where $a$ and $b$ are given by the $\displaystyle F_{\chi}(a) = \frac{\alpha}{2}$ and $\displaystyle F_{\chi}(b) = 1 - \frac{\alpha}{2}$ where $F_{\chi}$ denotes the cumulative distribution function.
A: It is described here how to construct confidence intervals for the variance in the normal model when $\mu$ is known (as well as when $\mu$ is unknown).
